The bisector of the

Given: ABC is a triangle; AD is the exterior bisector of ∠A and meets BC produced at D; BA is produced to F.
To prove:                 
Construction: Draw CE||DA to meet AB at E.
Proof: In Δ ABC, CE || AD cut by AC.
  ∠CAD = ∠ACE (Alternate angles)
Similarly CE || AD cut by AB
  ∠FAD = ∠AEC (corresponding angles)
Since ∠FAD = ∠CAD (given)
     ∴ ∠ACE = ∠AEC
 AC = AE ( by isosceles Δ theorem)
Now in Δ BAD, CE || DA

(Basic proportionality theorem):If a line is drawn parallel to one side of a triangle intersecting other two sides,
then it divides the two sides in the same ratio. '

By BPT,




But AC = AE (proved above) 





 

or (proved).